Question

a ball is dropped from a height of 2.50 m above the floor. a) find the speed v with which it reaches the floor. b) the ball now rebounds. the speed of the ball is decreased to 3v/4 due to this collision. how high will the ball rise?

Answer 1

Vo = 0 m/s
h = 2.5 m
g = 9.8 m/s^2

a. Vt^2 = Vo^2 + 2gh
v^2 = 0 + 2 × 9.8 × 2.5
v^2 = 49
v = √49
v = 7 m/s

b. The new speed v'
= 3v/4
= (3 × 7)/4
= 5.25 m/s

Vt^2 = v'^2 - 2gh
0 = 5.25^2 - 2 . 9.8 . h
0 = 27.5625 - 19.6h
19.6h = 27.5625
h = 27.5625 : 19.6
h = 1.40625 m

Answer 2

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[➊]

Vo ➙ 0 m/s
H ➙ 2.5 m
g ➙ 9.8 m/s^2

V ➙ √(2gH)
...➙ √(2 × 9.8 × 2.5)
...➙ 7 m/s

[➋]

(Kɪɴᴇᴛɪᴄ ᴇɴᴇʀɢʏ ᴏғ ʙᴀʟʟ ᴀғᴛᴇʀ ʟᴏꜱɪɴɢ ᴛʜᴇ ꜱᴏᴍᴇ ᴇɴᴇʀɢʏ ᴅᴜᴇ ᴛᴏ ᴄᴏʟʟɪꜱɪᴏɴ ᴀʀᴇ)

K➙ 0.5 mv^2

..➙ 0.5m × (3√/4)^2

..➙ 0.5m × (3 × 7/4)^2

..➙ 0.5m × (21 /4 )^2

(Tʜᴇɴ, Tʜɪꜱ ᴇɴᴇʀɢʏ ɪꜱ ᴄᴏɴᴠᴇʀᴛᴇᴅ ɪɴᴛᴏ ᴘᴏᴛᴇɴᴛɪᴀʟ ᴇɴᴇʀɢʏ ᴀᴛ ʜᴇɪɢʜᴛ 'ʜ')

....➙ 0.5m × (21 /4 )^2 = mgh

H ➙ (21 /4 )^2 / (2g)

H ➙ (21 /4 )^2 / (2 × 9.8)

H ➙ 1.40 m...✔

Sᴏ, ᴛʜᴇ ʙᴀʟʟ ᴡɪʟʟ ʀɪꜱᴇ ᴛᴏ ᴛʜᴇ ʜᴇɪɢʜᴛ ɪꜱ ᴇϙᴜᴀʟ ᴛᴏ 1.40m.